1 00:00:00,000 --> 00:00:08,820 PROFESSOR: Welcome back to recitation. 2 00:00:08,820 --> 00:00:10,850 In this video, what I'd like us to do 3 00:00:10,850 --> 00:00:13,400 is answer the following question. 4 00:00:13,400 --> 00:00:16,730 Suppose that f is a continuous, differentiable function. 5 00:00:16,730 --> 00:00:19,930 And if it's derivative, if f prime is never 0, 6 00:00:19,930 --> 00:00:23,010 and a is not equal to b, then show that f of a 7 00:00:23,010 --> 00:00:24,640 is not equal to f of b. 8 00:00:24,640 --> 00:00:26,640 I'm going to let you think about it for a while, 9 00:00:26,640 --> 00:00:28,490 see if you can come up with a good reason for that, 10 00:00:28,490 --> 00:00:30,323 and then I'll be back to explain my reasons. 11 00:00:38,010 --> 00:00:38,510 OK. 12 00:00:38,510 --> 00:00:42,300 Our object, again, is to show, if f 13 00:00:42,300 --> 00:00:46,290 is continuous and differentiable and its derivative is never 0 14 00:00:46,290 --> 00:00:49,200 and you're looking at two x-values that are different, 15 00:00:49,200 --> 00:00:51,217 show that their y-values have to be different. 16 00:00:51,217 --> 00:00:52,800 Show that if the inputs are different, 17 00:00:52,800 --> 00:00:54,256 the outputs have to be different. 18 00:00:54,256 --> 00:00:55,880 Now, this might remind you of something 19 00:00:55,880 --> 00:00:58,882 you saw in lecture about if the derivative has a sign, 20 00:00:58,882 --> 00:01:01,396 show the function-- if the derivative is positive, 21 00:01:01,396 --> 00:01:03,020 show the function is always increasing, 22 00:01:03,020 --> 00:01:04,394 or if the derivative is negative, 23 00:01:04,394 --> 00:01:06,160 show the function is always decreasing. 24 00:01:06,160 --> 00:01:10,340 So this is a similar type of problem to that. 25 00:01:10,340 --> 00:01:13,010 So what we're going to use is actually the mean value 26 00:01:13,010 --> 00:01:13,880 theorem. 27 00:01:13,880 --> 00:01:15,690 If you'll notice, I have f. 28 00:01:15,690 --> 00:01:18,380 It does satisfy the mean value theorem 29 00:01:18,380 --> 00:01:20,220 on an interval from a to b. 30 00:01:20,220 --> 00:01:22,420 I haven't even specified which is bigger, a or b. 31 00:01:22,420 --> 00:01:25,350 But it doesn't matter in this case. 32 00:01:25,350 --> 00:01:26,280 So what do we know? 33 00:01:26,280 --> 00:01:31,290 The mean value theorem tells us that if we look at-- well, 34 00:01:31,290 --> 00:01:37,990 let's just write it out-- f of b minus f of a over b minus a 35 00:01:37,990 --> 00:01:40,830 is equal to f prime of c-- and what 36 00:01:40,830 --> 00:01:43,200 do we know-- for c between a and b. 37 00:01:46,100 --> 00:01:49,300 So we want to know whether or not f of b minus f of a 38 00:01:49,300 --> 00:01:50,320 can ever be 0. 39 00:01:50,320 --> 00:01:52,740 We're trying to show that it cannot be 0. 40 00:01:52,740 --> 00:01:57,960 So we're going to isolate this expression and show that this 41 00:01:57,960 --> 00:01:59,530 subtraction cannot be 0. 42 00:01:59,530 --> 00:02:00,530 Well, how do we do that? 43 00:02:00,530 --> 00:02:03,427 Let me come over here to give us a little more room. 44 00:02:03,427 --> 00:02:05,260 I'm going to rewrite the mean value theorem. 45 00:02:05,260 --> 00:02:07,331 I'm going to multiply through by b minus a. 46 00:02:12,250 --> 00:02:18,460 So we get f prime of c time b minus a. 47 00:02:18,460 --> 00:02:20,090 Now, we just want to show, again, 48 00:02:20,090 --> 00:02:22,530 that f of b minus f of a cannot be 0. 49 00:02:22,530 --> 00:02:24,570 What's the only thing-- well, not only thing, 50 00:02:24,570 --> 00:02:26,570 we know two things-- what two things do we know? 51 00:02:26,570 --> 00:02:29,350 We know f prime of c is not 0. 52 00:02:29,350 --> 00:02:30,380 That was given to you. 53 00:02:30,380 --> 00:02:33,360 f prime is never 0, so certainly at any fixed value, 54 00:02:33,360 --> 00:02:35,360 f prime of c is not 0. 55 00:02:35,360 --> 00:02:37,860 So we know this term is not 0. 56 00:02:37,860 --> 00:02:39,720 We also know that b is not equal to a, 57 00:02:39,720 --> 00:02:42,160 so we know b minus a is not 0. 58 00:02:42,160 --> 00:02:45,170 The only way to get a product of two numbers to be 0 59 00:02:45,170 --> 00:02:46,380 is if one of them is 0. 60 00:02:46,380 --> 00:02:51,007 So this in fact, this product is not equal to 0. 61 00:02:51,007 --> 00:02:52,840 The fact that this product is not equal to 0 62 00:02:52,840 --> 00:02:55,950 tells us f of b minus f of a is not equal to 0. 63 00:02:55,950 --> 00:02:59,600 And that alone is enough to conclude that f of b 64 00:02:59,600 --> 00:03:03,024 is not equal to f of a. 65 00:03:03,024 --> 00:03:04,690 So, again, let me just point out of this 66 00:03:04,690 --> 00:03:08,140 is probably reminds you very much of the type of thing 67 00:03:08,140 --> 00:03:11,860 you've seen where you were showing if f prime had a sign, 68 00:03:11,860 --> 00:03:15,400 then you could determine whether f was increasing or decreasing. 69 00:03:15,400 --> 00:03:17,200 It's the same type of problem as that. 70 00:03:17,200 --> 00:03:20,180 It's exploiting what the mean value theorem tells you. 71 00:03:20,180 --> 00:03:21,835 So I think we'll stop there.